Mathematics - 2020-09-15

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Mike Miller

00:00

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he's gotta be fucking with me

user2103480

I think this is not even close to the worst passage in that article

Mike Miller

Look man I get how hard it is to make good notation, I've made plenty of bad and overloaded notstion

But cmon

Ted Shifrin

@Jeff What you are saying is all wrong. If you want a rigorous argument for what I suggested, remember that $$\lim(f(x)-15) = \lim\left(\frac{f(x)-15}{x-4}\cdot (x-4)\right) = \lim\frac{f(x)-15}{x-4}\cdot\lim(x-4).$$

user2103480

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@MikeMiller ah yes, straightforward mochizuki style. but he probably got those 3m for a reason

I'm taking SPDE next semester and have not a single clue how anything he talks about is even slightly connected to that, after like 4 consecutive lectures on probability

Jeff

00:18

@TedShifrin Where do I go from here than? I guess I don't understand what you said last night.

Ted Shifrin

@Jeff: Did you work out what I typed there? What did it tell you?

Jeff

I see it. I'm not sure what it's telling me. We know that the left limit = 3 and the lim(x-4)=0

@TedShifrin

Ted Shifrin

So what's the product of those limits?

Jeff

zero

Ted Shifrin

So what does that say $\lim (f(x)-15)$ is ?

Jeff

00:20

zero?

Ted Shifrin

Why are you putting the ?

Jeff

not sure. why. but lim(x-4)/lim(x-4) =1, so lim(f(x)-15) must be zero

Thorgott

that above screenshot unironically looks like something out of the random math paper generator

Ted Shifrin

The first equality in what I typed used the fact that $\dfrac{x-4}{x-4}=1$ whenever $x\ne 4$, yes.

No limits there.

Jeff

@Ted TY

Ted Shifrin

00:23

So, what I said last night was that you should "know" that if $a/b$ has a limit and $b\to 0$, then $a$ must go to $0$ as well. This algebra is the rigorous proof.

Mike Miller

@user2103480 I imagine it's clear why he wants to do this to experts in SPDE who will understand the failed attempts at doing what he succeeds at doing

Unfortunately math papers tend to not be the best place to learn math

Ted Shifrin

LOL, yes, that's true. Mostly journals try to shorten articles and therefore minimize exposition and explanation. But I fooled a few :P

4 hours later…

Balarka Sen

04:36

@AlessandroCodenotti Hahaha lmao

I thought somehow you were doing C^* coalgebras

xcodeking

Similar to how one can write an integral in terms of a power series, is there a standard way to convert a power series to an integral?

Think of, say, the Fresnel integral, but backwards

Sayan Chattopadhyay

04:57

@xcodeking What do you mean by integral in terms of power series specifically? That is say if you're calculating $\int f dx$, and there is some open neighborhood around a point in the domain of $f$ such that $f = \sum_n a_nx^n$, then you have $\int f dx = \int \sum_n a_nx^n$
Writing this as a power series would mean you have to interchange the integral and the summation, so you have to have (I think) some bound on the partial sums and then you can interchange by DCT. Is this what you have in mind?

2 hours later…

Ramanewbie

07:06

@TedShifrin Yay ! :D Well I've been quite busy for the last few years. Partly doing maths away from this chat ;)
In fact I'm more into computer science and electronics nowadays
What about you, anything new ?

2 hours later…

Alessandro Codenotti

08:57

@BalarkaSen oh god no

Hawk

09:19

If N = 0, then what does it mean to have a free rank of 0? I thought N = 0 means that it ISN'T free.

Leaky Nun

09:32

zero module

Hawk

It has no basis, so what does free here mean?

Leaky Nun

the basis is the empty set

"the empty set is a basis" is different from "it has no basis"

2 hours later…

skullpatrol

11:11

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Adam L

12:10

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yh05

12:26

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These are different definitions of limit at infinity, first from Hunter's intro real analysis book, second from Bartle's intro real analysis book, and third from baby Rudin.

Is there a standard definition of limit at infinity? Which definition should I adopt?

Thorgott

Second definition is a special case of the first definition and first definition is a special case of the third definition

They're all the standard definition, just on different levels of generality

yh05

12:46

@Thorgott Does the third definition (from Baby Rudin) coincides with Definition 4.1 when A and x are real? (Rudin claimed that it does in his book).

(Definition 4.1) $\quad$ Let $X$ and $Y$ be metric space; suppose $E \subset X$, $f$ maps $E$ into $Y$, and $p$ is a limit point of $E$. We write $f(x)\to q$ as $x\to p$, or $\lim_{x\to p}f(x)=q$ if there is a point $q \in Y$ with the following property: For every $\epsilon > 0$ there exists a $\delta > 0$ such that $d_Y(f(x),q)<\epsilon$ for all points $x\in E$ for which $0<d_X(x,p)<\delta$.

Mike Miller

I haven't even read the definitions here but I see no reason to doubt Rudin on this --- try proving the equivalence

Thorgott

Yes, it does

yh05

@Thorgott I thought it doesn't coincide with definition 4.1. I came up with an example, where did I go wrong?

Let E = {0,1}. f is defined on E with f(0) = f(1) = 2. Then according to definition 4.33, f(t) -> 2 as t -> 0, since for every neighbourhood of 2, I can take the deleted neighbourhood of 0 with radius 2 (which is {1}).

But 0 is not a limit point of E.

Tobias Kildetoft

13:02

@yh05 You need an actual neighborhood of the point according to that definition. The only neighborhoods of $0$ are $\{0\}$ and $E$.

Thorgott

ah

true, the definitions only coincide when $x$ is a limit point, because the other definition isn't concerned with that case

robjohn

@skullpatrol I've found that gas often helps with social distancing

Thorgott

at second look, there seems to be an error in Rudins definition

$V$ should be a deleted neighborhood of $x$

yh05

So I need to add the assumption that x is a limit point to definition 4.33 when x is real?

@TobiasKildetoft I chose the neighborhood (-2,2) in my example.

Tobias Kildetoft

@yh05 as which one?

Thorgott

13:13

I mean, you don't have to assume $x$ is a limit point for the definition to make sense, though I don't think there's ever any reason to consider the other cases

But I think you do have to assume $V$ to be a deleted neighborhood of $x$ to get anything sensible

Otherwise limits at isolated points are non-unique by that definition and that's generally undesirable

Tobias Kildetoft

Ahh, indeed the conclusion that the limit at 0 is 2 is correct

It is counter-intuitive, but it makes more sense if you think of extending the function to be equal to 2 everywhere but at 0

And then it also becomes clear that it should indeed be a deleted neighborhood

Mike Miller

my bad for the blind defense of Rudin!

yh05

I think Rudin mean deleted neighborhood for the case that x is real in the definition. At the end of the definition it reads for all $t\in V\cap E, t\ne x$

Tobias Kildetoft

Ahh, that does make it all fit together then

Rudin should just have ordered the quantifiers correctly and it would have been much clearer

Thorgott

Yeah, the important bit is that you want $V\cap E\setminus\{x\}$ to be non-empty and not just $V\cap E$ to be non-empty, cause if $V\cap E=\{x\}$, then the latter condition is vacuously satisfied and limits aren't unique anymore

Tobias Kildetoft

13:20

True

clearly we should write a replacement for Rudin, because I for one feel up to doing it better :)

yh05

My understanding so far, definitions 4.1 and 4.33 do not coincide when A and x are real. If I pre-suppose that x is a limit point of E, then the definitions coincide. But it is necessary for x to be a limit point of E since limit is not well defined if x is an isolation point of E. So in the end, definition 4.33 is fine?

Thorgott

Definition $4.33$ should be changed as to demand $V$ to be a deleted neighborhood of $x$. It might very well be the case that changing this definition does not affect the correctness of anything else in the book, but it is definitely the more mathematically sensible definition.
Everything is well-defined in any case, you just don't get uniqueness with the original definition 4.33 in case $x$ is isolated. It's not necessary for $x$ to be a limit point of $E$ for definition 4.33 to make sense, but I'm like 99% sure you will never need to consider a limit at a point that's not a limit point.

Perhaps the better way to put this is that definition 4.33 subsumes definition 4.1

geocalc33

anyone know how to make an image in the question body smaller?

yh05

I will change neighborhood V of x to deleted neighborhood V of x in my book. Thank you for your help.

S. Das

13:40

This is supposed to be something basic. But I can't figure out the mistake in the reasoning. Let $(X,\le)$ be a poset such that every chain has an upper bound.

Let $\Sigma$ be a (non-empty) chain of $X$ and consider the set of all its upper bounds. Let's denote it by $\mathsf{UpBound}(\Sigma)$. Consider any chain of $\mathsf{UpBound}(\Sigma)$. Since by hypothesis, every chain has an upper bound, this chain much have an upper bound as well, say $x$. Now we claim that $x$ is the maximal element of $X$. To show this observe that if $y\in X$ be such that $x<y$ then $y\in \mathsf{UpBound}(\Sigm…

Question: Where do we implicitly use Axiom of Choice in the argument?

Thorgott

$x$ is not an upper bound of $\operatorname{UpBound}(\Sigma)$, it's an upper bound of some non-specified chain in $\operatorname{UpBound}(\Sigma)$

S. Das

Oh man!

Thanks @Thorgott.

Thorgott

np

geocalc33

14:00

I need a hint on how to rotate $z^2-y^2-x^2=1$ so that the forward sheet is in the first octant $(+,+,+)$

yh05

One question, which definition of limit at infinity (the ones I posted above) is more commonly used?

Thorgott

They all agree wherever they apply

an author will just choose whichever formulation/level of generality they feel appropriate for what they're trying to achieve

2 hours later…

user330477

15:55

Show that for $N$ large enough, there exist $a,b \in \mathbb{N}$ such that $N<a^2+b^2<N+N^{2/3}$. Any hints on this one would be appreciated.

3 hours later…

user10478

18:30

Stupid question, if $y$ is really a function of $x$, but we want to find $\frac{dx}{dy}$, is it sufficient to solve for $x$ and follow differentiation rules as though $x$ were a function and $y$ an independent variable, or do you have to do something in reverse when applying, i.e., the chain rule?

My understanding is that, at least when the relation between $x$ and $y$ is both right-unique and left-unique, the roles of dependent and independent variable are defined by what derivative is being taken, rather than being a property of the expression or equation itself, so that $\frac{dx}{dy}$ defines $y$ as the independent variable for that derivative, but I wanted to make sure this is correct.

1 hour later…

Ted Shifrin

19:40

@user10478 It's a theorem (Inverse Function Theorem) that if $\frac{dy}{dx}$ evaluated at the point $(x_0,y_0)$ is nonzero, then locally you can solve for $x=x(y)$ near that same point (even if you cannot do so explicitly), and then $\frac{dx}{dy}$ at that point is the reciprocal. So, yes, locally you can make $y$ the independent variable. There may be lots of different $x$ values for that $y$, but we're staying near the point $(x_0,y_0)$.

2 hours later…

Charlie

21:10

Is there a name for functions that can be recovered by taking their derivative repeatedly? sine and cosine functions for example

Leaky Nun

Analytic function

In mathematics, an analytic function is a function that is locally given by a convergent power series. There exist both real analytic functions and complex analytic functions, categories that are similar in some ways, but different in others. Functions of each type are infinitely differentiable, but complex analytic functions exhibit properties that do not hold generally for real analytic functions. A function is analytic if and only if its Taylor series about x0 converges to the function in some neighborhood for every x0 in its domain. == Definitions == Formally, a function...

Charlie

Ah I mean as in $f^{(n)}(x)=f(x)$ for instance

for some n

Leaky Nun

then I don't know if there's a name for that

Ted Shifrin

I know no name.

If you fix any $n$, you can write down the power series for any such function.

Mike Miller

They're linear combinations of $e^{\omega x}$ where $\omega$ is an $n$th root of unity

Leaky Nun

21:25

aha, because it's a linear differential equation

Yahyaa

21:59

Hello

This looks new

Hmm ....

skullpatrol

22:47

Welcome

@robjohn but sir, green house gas is affecting the ozone layer?

In other news:

India confirms now 93,000 cases per day. But isn’t testing widely. The true number of new infections is estimated to be up to 800,000 per day.

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